Proof

given by postcomposition with ff (the dependent sum along ff). Therefore by prop. 1 it is sufficient to show that for all (A→X2)(A \to X_2) in 𝒞/X2\mathcal{C}_{/X_2} and (B→bX1)∈𝒞/X1(B \stackrel{b}{\to} X_1) \in \mathcal{C}_{/X_1} that

B×X1f*A≃B×X2A
B \times_{X_1} f^* A \simeq B \times_{X_2} A

in 𝒞\mathcal{C}. But this is the pasting law for pullbacks in 𝒞\mathcal{C}, which says that the two consecutive pullbacks on the left of